3
$\begingroup$

I should preface this question by saying that I strongly suspect the answer is negative, but I couldn't find the counterexample myself.

Say we are working on the unit disc $D \subset \mathbf{R}^n$, where we are given two uniformly elliptic operators with coefficients $A^{ij}$ and $a^{ij}$. (These may be as regular as one needed, for instance at least $C^1$. Moreover for definiteness one might be the identity matrix, say $a^{ij} = \delta^{ij}$.)

Let $U$ and $u$ be two solutions respectively of the divergence-form equations $D_i(A^{ij} D_j U) = 0$ and $D_i(a^{ij} D_j u) = 0$. (Again $U$ and $u$ can be taken as regular as one desires.) We consider their difference $v:= U - u$.

Question. Is there a function $\lambda: D \to [0,1]$ so that $v$ satisfies the 'intermediate' equation $D_i((\lambda A^{ij} + (1 - \lambda) a^{ij}) D_j v) = 0$?

A negative answer would follow for example from the failure of $v$ to satisfy some property of elliptic PDE, the maximum principle being the obvious candidate. As mentioned above, I couldn't find the example that demonstrates this failure. (I suspect there might even exist one where $A^{ij}$ and $a^{ij}$ are constant.)

$\endgroup$
3
  • 3
    $\begingroup$ $U = 2x^2-y^2$ and $u = x^2-2y^2$ solve constant-coefficient elliptic equations, but $U-u = x^2+y^2$ has an interior minimum and thus cannot solve an elliptic equation. $\endgroup$ Commented Jun 30, 2021 at 15:45
  • $\begingroup$ @ConnorMooney Thanks! Would you like to post this as answer? $\endgroup$
    – Leo Moos
    Commented Jun 30, 2021 at 16:29
  • $\begingroup$ Sure, thank you $\endgroup$ Commented Jun 30, 2021 at 16:50

1 Answer 1

4
$\begingroup$

$U = 2x^2-y^2$ and $u = x^2-2y^2$ solve constant-coefficient elliptic equations, but $U - u = x^2 + y^2$ has an interior minimum and thus cannot solve an elliptic equation.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .